Tính I =Lim căn{16{x^2} + 5x} / 2x - 1

Chọn B

Ta có \[I = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {16{x^2} + 5x} }}{{2x - 1}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {{x^2}\left( {16 + \frac{5}{x}} \right)} }}{{x\left( {2 - \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - x\sqrt {16 + \frac{5}{x}} }}{{x\left( {2 - \frac{1}{x}} \right)}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{\sqrt {16 + \frac{5}{x}} }}{{2 - \frac{1}{x}}} = 2\].